A contractor employs 15 men to complete a work in 24 days, working 8 hours per day. After six days, he adds five more men but also reduces the working hours to six per day from the seventh day onwards. If all men work with the same efficiency throughout, how many more days will be required to finish the remaining work?

Correct — B. Measure the job in man-hours.

Whole job = 15 × 24 × 8 = 2,880 man-hours
Done in 6 days = 15 × 6 × 8 = 720
Left = 2,880 − 720 = 2,160

New daily rate = 20 men × 6 hours = 120 man-hours
Days needed = 2,160 ÷ 120 = 18 → option (b)

Why the others are wrong

• (a)
  20 — 20 days at 120 man-hours a day is 2,400 man-hours, 240 more than the 2,160 left after the first six days.
• (c)
  21 — 21 days at 120 man-hours a day is 2,520 man-hours, 360 more than the 2,160 left.
• (d)
  22 — 22 days at 120 man-hours a day is 2,640 man-hours, 480 more than the 2,160 left.

With every worker equally efficient, a job is a fixed stock of man-hours (men × days × hours a day), and each day removes men × hours of it.

Here the change is neutral: 20 men × 6 hours = 120 = 15 men × 8 hours. The daily output is the same, so the 18 days left in the original plan stay 18, and the job still ends on day 24.

Key facts

• M₁ × D₁ × H₁ = M₂ × D₂ × H₂ for the same work at equal efficiency.
• Work left = total man-hours − man-hours already worked.
• If men × hours a day is unchanged, the days still needed are unchanged.

Common traps

• Answering the total, 6 + 18 = 24 days: the question asks how many more days.
• Scaling only by men and forgetting the cut to 6 hours: 2,160 ÷ 160 = 13.5 days.
• Scaling only by hours and forgetting the five new men: 2,160 ÷ 90 = 24 days.

No directly related past PYQ was found.